Single-slit diffraction appears when light passes through a narrow opening whose width is not extremely large compared with the wavelength. Instead of continuing as a perfectly sharp beam, the light spreads out and produces a characteristic intensity envelope on a distant screen. The pattern contains one broad bright central maximum and weaker side maxima separated by dark minima.
The standard explanation comes from interference between wavelets emitted from different parts of the slit. Every point across the slit acts as a secondary source. At some observation angles, the contributions from different parts of the slit reinforce each other strongly. At other angles, they cancel. This is why diffraction is really an interference effect produced by a continuous distribution of sources across the slit width.
The most important condition is the location of the dark minima. For a slit of width \(a\), the minima occur when
\[
a\sin\theta_m = m\lambda,
\qquad
m=\pm1,\pm2,\pm3,\dots
\]
where \(\lambda\) is the wavelength and \(\theta_m\) is the angle of the \(m\)-th minimum measured from the central axis. When the angle is small, we often use the approximation \(\sin\theta \approx \theta\) in radians, so
\[
\theta_m \approx \frac{m\lambda}{a}.
\]
For the sample input
\(a=0.1\ \text{mm}\)
and
\(\lambda=600\ \text{nm}\),
convert to SI units:
\[
a=0.1\times10^{-3}\ \text{m}=1.0\times10^{-4}\ \text{m},
\qquad
\lambda=600\times10^{-9}\ \text{m}=6.0\times10^{-7}\ \text{m}.
\]
Then the first minimum is
\[
\theta_1 \approx \frac{\lambda}{a}
= \frac{6.0\times10^{-7}}{1.0\times10^{-4}}
= 6.0\times10^{-3}\ \text{rad}.
\]
So the first minimum is at about
\[
\theta_1 \approx 0.006\ \text{rad},
\]
exactly as in the sample result.
The angular width of the central maximum is determined by the distance between the first minimum on one side and the first minimum on the other side. Using the small-angle approximation, this gives
\[
\Delta\theta_{\text{central}} \approx 2\frac{\lambda}{a}.
\]
This formula is important because it shows that a narrower slit produces a wider diffraction pattern. That may look surprising at first, but it is a central result of wave optics: tighter confinement in the slit gives stronger spreading afterward.
To find the intensity at any observation angle, we use the single-slit envelope formula
\[
I = I_0\left(\frac{\sin\beta}{\beta}\right)^2,
\qquad
\beta=\frac{\pi a \sin\theta}{\lambda}.
\]
Here \(I_0\) is the central maximum intensity scale. When \(\theta=0\), we have \(\beta=0\). The expression \(\sin\beta/\beta\) then approaches 1 in the limit, so the center of the pattern is the brightest point.
As \(\theta\) moves toward a minimum, the value of \(\beta\) approaches \(m\pi\), so \(\sin\beta\) goes to zero and the intensity drops to zero. Between these minima, the function rises again to form weaker secondary maxima.
The animation in this calculator shows incoming plane waves, the slit opening, and representative diffracted rays heading toward a chosen probe angle. The screen strip and the graph both represent the same idea: the intensity is largest at the center, decreases away from the center, and reaches exact zeros at the minima.
At a more advanced university level, one studies Fresnel diffraction, finite screen distances, and the connection between diffraction and Fourier transforms. But for the far-field or Fraunhofer case, the formulas above capture the essential behavior very well.
So the key relations to remember are
\[
a\sin\theta_m = m\lambda,
\qquad
\beta=\frac{\pi a \sin\theta}{\lambda},
\qquad
I = I_0\left(\frac{\sin\beta}{\beta}\right)^2,
\qquad
\Delta\theta_{\text{central}} \approx 2\frac{\lambda}{a}.
\]
These formulas tell you where the dark fringes occur, how wide the central maximum is, and how the full diffraction envelope varies with angle.
